Finding the Closest Point on a Parabola: How Can We Minimize Distance?

[fubag]

Find the point P on the parabola y = x^2 closest to the point (1,0).

To solve for P...

I used the distance formula, and then took the derivate. Using points (x, x^2) and (1,0)...

however, while taking the derivative I get a really nasty 4th power equation, which I am not sure I can solve without a calculator.

 

[EnumaElish]

Can you derive your solution here?

 

[DeadWolfe]

I don't know how you did it, but the big hint here is of course to minimize the distance *squared*, rather than the distance.

 

[fubag]

ok this is what I have so far...

D = sqrt. ((x-2)^2 + (x^2)^2)

then I took the derivative...dD/dx = (1/2)((x-1)^2 + x^4)^-1/2)*(2(x-1) + 4x^3)

this simplifies to..

dD/dx = ((x-1) +2x^3)/sqrt.((x-1)^2 + x^4)

setting that equal to zero...

this is where I encounter issues..

2x^3 + x - 1 =0

and

sqrt. ((x-1)^2 + x^4) = 0

 

[HallsofIvy]

As Dead Wolfe suggested, try taking the derivative of D² instead.